I want to prove that the mapping of the complex plane to the complex plane is a closed map.
Only way I can think of is to prove first that it is a proper map.
Is there a more straightforward way ?

June 27th 2011, 10:09 PM

Jose27

Re: How do you prove that z->z^2 is a closed map ?

It's obvious that is closed, and since it leaves fixed it behaves nice with respect to unbounded closed sets (an unbounded set is closed in the plane iff the set plus is closed in the sphere). ( denotes the Riemann sphere)

June 29th 2011, 07:07 PM

Aki

Re: How do you prove that z->z^2 is a closed map ?

Thank you for your help.

From your reply I got the impression that the necessary and sufficient condition for a continuous mapping of into to be a closed map is to leave fixed.
Am I right ?

June 29th 2011, 08:15 PM

Jose27

Re: How do you prove that z->z^2 is a closed map ?

Quote:

Originally Posted by Aki

Thank you for your help.

From your reply I got the impression that the necessary and sufficient condition for a continuous mapping of into to be a closed map is to leave fixed.
Am I right ?

Not at all: Take where the natural projection on the real axis, then fixes (in the sense that there is a mapping of the sphere to the sphere that coincides with via stereographic projection, ie. as usual), but it's not difficult to see that is not closed. On the other hand if is a constant, we can extend to the sphere and this is, trivially, a closed continous map that doesn't fix .

What I meant to say was that obviously your mapping takes compact sets to compact sets (both in the plane), so we only need to check whether it sends unbounded closed sets (say ) to closed sets ( say), but by adding to we get a closed set in the sphere and thus is closed in the sphere, but by the first post is closed in the plane, but this is exactly and we're done.